use DeMoivre's theorem to simplify ?

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use DeMoivre's theorem to simplify ?

How do you use DeMoivre's theorem to simplify ${(- {(3)}^{0.5} + i)}^{\frac{9}{2}}$ $(-(3)^0.5+i)^(9/2)$

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Answer 1 · 2018-01-17T10:23:23

${(- \sqrt{3} + i)}^{\frac{9}{2}} = 16 - 16 i$ $(-sqrt3+i)^(9/2)=16-16i$

Explanation:

DeMoivre's theorem states that if a complex number in polar form is $a + i b = r (cos θ + i sin θ)$ $a+ib=r(costheta+isintheta)$ , then

${(a + i b)}^{n} = r^{n} (cos n θ + i sin n θ)$ $(a+ib)^n=r^n(cosntheta+isinntheta)$

Now we can write ${(- {(3)}^{0.5} + i)}^{\frac{9}{2}}$ $(-(3)^0.5+i)^(9/2)$ as

${(- \sqrt{3} + i)}^{\frac{9}{2}}$ $(-sqrt3+i)^(9/2)$

= ${2 (- \frac{\sqrt{3}}{2} + i \frac{1}{2})}^{\frac{9}{2}}$ ${2(-sqrt3/2+i1/2)}^(9/2)$

= ${2 (cos (\frac{5 π}{6}) + i sin (\frac{5 π}{6}))}^{\frac{9}{2}}$ ${2(cos((5pi)/6)+isin((5pi)/6))}^(9/2)$

= $2^{\frac{9}{2}} (cos (\frac{5 π}{6} \times \frac{9}{2}) + i sin (\frac{5 π}{6} \times \frac{9}{2})}$ $2^(9/2)(cos((5pi)/6xx9/2)+isin((5pi)/6xx9/2)}$

= $16 \sqrt{2} (cos (\frac{15 π}{4}) + i sin (\frac{15 π}{4}))$ $16sqrt2(cos((15pi)/4)+isin((15pi)/4))$

= $16 \sqrt{2} (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))$ $16sqrt2(cos(-pi/4)+isin(-pi/4))$

= $16 \sqrt{2} (\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}})$ $16sqrt2(1/sqrt2-i1/sqrt2)$

= $16 - 16 i$ $16-16i$

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use DeMoivre's theorem to simplify ?

How do you use DeMoivre's theorem to simplify ${(- {(3)}^{0.5} + i)}^{\frac{9}{2}}$ $(-(3)^0.5+i)^(9/2)$

1 Answer

Explanation:

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use DeMoivre's theorem to simplify ?

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How do you use DeMoivre's theorem to simplify ${(- {(3)}^{0.5} + i)}^{\frac{9}{2}}$ $(-(3)^0.5+i)^(9/2)$